91Ó°ÊÓ

Chemical reactions often reach a point where the rate of the forward reaction equals the rate of the backward reaction. At this point, known as equilibrium, the concentrations of the reactants and products remain constant.
This balance is quantified by the equilibrium constant, represented as \(K_c\).
The equilibrium constant for a reaction is derived from the concentrations of the reactants and products at equilibrium. For instance, in the reaction \(2 \mathrm{H}_2(g) + \mathrm{S}_2(g) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{S}(g)\), the expression for \(K_c\) is:
\[K_c = \frac{[\mathrm{H}_2 \mathrm{S}]^2}{[\mathrm{H}_2]^2[\mathrm{S}_2]}\]

If \(K_c\) is much greater than 1, the position of equilibrium lies towards the products, meaning the reaction favors the formation of products.
Conversely, if \(K_c\) is much less than 1, the reaction favors the reactants. In the provided exercise, \(K_c = 1.08 \times 10^7\) at \(700^{\circ}\)C is very large, indicating a significant amount of \(\mathrm{H}_2 \mathrm{S}\) compared to the reactants.

When dealing with reactions involving gases, the equilibrium constant can also be expressed in terms of partial pressures, represented by \(K_p\).
Like \(K_c\), \(K_p\) provides essential insights into the position of equilibrium.
To convert \(K_c\) to \(K_p\), we use the formula: \[K_p = K_c(RT)^{\Delta n}\]
where:

• \(R\) is the ideal gas constant, approximately 0.0821 atm L/mol K
• \(T\) is the temperature in Kelvin
• \(\Delta n\) is the change in moles of gas (products minus reactants)

In the provided exercise, we calculated \(\Delta n\) as \(-1\) since the number of moles on the product side is less than that on the reactant side.
By substituting the values, we obtained \(K_p = 2.68 \times 10^8\), indicating that at 700°C, the equilibrium mixture highly favors the product formation as well.

Stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction.
It is essential for understanding how changes in reaction conditions affect equilibrium constants. In the rewritten reaction \(\mathrm{H}_2(g) + \frac{1}{2} \mathrm{S}_2(g) \rightleftharpoons \mathrm{H}_2 \mathrm{S}(g)\), the stoichiometry changes because the coefficients are altered.
When the stoichiometric coefficients are modified, it affects the equilibrium constant calculation.
In this exercise, changing the coefficients to half their original value requires taking the square root of the initial \(K_c\), resulting in a new \(K_c\) of \(3.286 \times 10^3\).
This adjustment helps us understand how even minor stoichiometric changes can impact the equilibrium constants and, consequently, the reaction dynamics.